SOLUTION: https://vle.mathswatch.co.uk/images/questions/question2399.png. please I need help with this hard problem :) thanks

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Question 1125383: https://vle.mathswatch.co.uk/images/questions/question2399.png. please I need help with this hard problem :) thanks
Found 2 solutions by MathLover1, greenestamps:
Answer by MathLover1(20849) About Me  (Show Source):
You can put this solution on YOUR website!

here you have two right angle triangles, CEA and CDB
the area o the trapezium ABDE is equal to the difference of the area of CEA and CDB
area_of_ABDE=area_of_CEA+-area_of_CDB

the area_of_CEA=%28%288cm%2BDEcm%29%2A9cm%29%2F2 ->as you can see, we need to find DE
since CEA and CDB are similar triangles, corresponding sides are proportional; so, we have
8%3A6=%288%2BDE%29%3A9 ...solve for+ED
8%2A9=%288%2BDE%29%2A6
72=48%2B6DE
72-48=6DE
24=6DE
DE=24%2F6
DE=4cm
and the area_+of_+CEA=%28%288cm%2B4cm%29%2A9cm%29%2F2+=%2812cm%2A9cm%29%2F2=6cm%2A9cm=54cm%5E2
the area_of_CDB=%288cm%2A6cm%29%2F2=4cm%2A6cm=24cm%5E2

finally,the area o the trapezium
area_of_ABDE=54cm%5E2-24cm%5E2=30cm%5E2


Answer by greenestamps(13196) About Me  (Show Source):
You can put this solution on YOUR website!


Triangles BCD and ACE are similar.

The ratio of BD:CD is 6:8 = 3:4; so the ratio of AE:CE must be 3:4.

Then, since AE=9, CE=12.

Then since you know CD=8, you know the length of DE, which is the altitude of the trapezium.

And then the area of the trapezium is the height times the average of the two bases.